Oxidation levels in organic chemistry — Oxidation & Reduction

Oxidation and reduction are defined in general chemistry by electron transfer, and that definition is correct and almost unusable here. Organic reactions rarely move a whole electron between two clearly identifiable ions; they change what a carbon is bonded to. So organic chemists use a working test that gives the same answers with far less arithmetic.

The working test

Oxidation = gaining bonds to oxygen (or another electronegative atom), or losing bonds to hydrogen.
Reduction = the reverse: losing bonds to oxygen, or gaining bonds to hydrogen.

That is enough for almost every reaction in this course. Turning an alcohol into a ketone replaces a C–H with a second C–O bond, so it is an oxidation. Turning a ketone into an alcohol does the reverse, so it is a reduction. Adding H₂ across a double bond adds two C–H bonds, so hydrogenation is a reduction — which is worth saying explicitly, because "hydrogenation" does not contain the word.

A reaction that changes neither count is neither. Adding HBr across an alkene gains a C–H and a C–Br; the hydrogen count went up and the electronegative-atom count went up, and those cancel. Hydration, likewise, adds one C–H and one C–O. Most of the additions from the alkene chapter are redox-neutral, and noticing that is a fast way to check you have not mislabeled something.

The carbon oxidation ladder

If you want the actual number, the rule for a carbon's oxidation state is:

Each bond to a more electronegative atom (O, N, halogen) counts +1.
Each bond to hydrogen counts −1.
Each bond to another carbon counts 0 — which is why you can usually ignore the rest of the molecule.

Run that over the one-carbon compounds and you get a ladder that organizes most of what follows:

Compound	Carbon oxidation state	Class
CH₄	−4	alkane
CH₃OH	−2	alcohol
CH₂O	0	aldehyde
HCO₂H	+2	carboxylic acid
CO₂	+4	fully oxidized

Each step up is a two-electron oxidation. Reading it as a ladder makes several later facts obvious rather than memorized: an alcohol and an alkyl halide sit at the same rung, so converting one to the other is not a redox reaction at all. A carboxylic acid, an ester, an amide and a nitrile also all sit at +3 together, which is why interconverting them is substitution rather than oxidation — the same point the nomenclature chapter made about the top of the priority list.

Why 3° alcohols cannot be oxidized

The ladder with its second dimension drawn in. A rung is not one compound but a whole set of them — everything whose carbon carries the same number of bonds to oxygen, nitrogen or halogen — and moving sideways along a rung costs no oxidant at all.This is why an alcohol and an alkyl halide interconvert with nothing more than a nucleophile, and why an ester, an amide and a nitrile interconvert with each other but never with an aldehyde. The oxidation-state column is the one-carbon case, so formic acid reads +2 here because that carbon still carries an H; put an alkyl group there instead and a carboxylic acid is +3. The rung is the durable fact, not the number.

Look at what oxidation of an alcohol actually does: it removes the hydrogen from the carbon bearing the OH — the carbinol carbon — and forms the second bond to oxygen in its place. The reaction needs a C–H bond at that position to take.

A primary alcohol has two hydrogens there, so it can be oxidized twice: to an aldehyde, then on to a carboxylic acid.
A secondary alcohol has one, so it goes once, to a ketone, and stops.
A tertiary alcohol has none, so it does not react at all under ordinary oxidizing conditions.

This is not a rule to memorize separately. Count the hydrogens on the carbinol carbon and the answer falls out, including the number of times the reaction can run.

The shorthand [O] and [H] in a scheme means "an oxidizing agent" and "a reducing agent" without committing to which one. It is useful when the point is the transformation rather than the reagent — but most exam questions are about which reagent, precisely because the choice controls where the reaction stops.

Selectivity is the whole subject

Almost nothing in the next four sections is about whether a transformation is possible. It is about stopping it in the right place and leaving the rest of the molecule alone.

Oxidizing a primary alcohol is the clearest case. The same substrate gives an aldehyde with one reagent and a carboxylic acid with another, and the difference is not how much you add or how long you wait — it is the choice of reagent and, in the chromium cases, whether water is present. Reducing a carbonyl is the mirror image: two common hydride reagents differ enormously in what they will and will not touch, and choosing the weaker one on purpose is how you reduce a ketone in a molecule that also contains an ester.

So as you read the next sections, keep two questions in front of the reagent list: where does this stop, and what else in the molecule does it attack. Those two answers are what a synthesis question is really asking for.

What carries forward

The ladder is the organizing idea for the rest of this chapter and reappears in synthesis planning, where "I need to go up two rungs and there is no single reagent for it" is a routine reason to use two steps. It also explains a fact from the carbonyl chapters that otherwise looks arbitrary: acyl derivatives interconvert freely because they are all at the same oxidation level, and getting off that rung — down to an aldehyde or an alcohol — needs a reducing agent rather than another nucleophile.